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Here cites its original claim from http://www.math.tau.ac.il/~ostrover/Teaching/18125.pdf.

Theorem 3.1 Jensen's Inequality

Let $(X,\mathcal{M},\mu)$ be a probability space (a measure space with $\mu(X) = 1)$, $f: X \to \mathbb R \in L^1(X, \mu)$, and $\psi: \mathbb R \to \mathbb R$ be a convex function, then $$\psi\int_X f d\mu \le \int_X (\psi \circ f)d\mu$$.

Proof:

Since $\psi$ is convex, at each $x_0 \in \mathbb R$, there exist $a,b \in \mathbb R$ such that $\psi(x_0) = ax_0 + b$ and $\psi(x) \ge ax + b, \forall x \in \mathbb R$, (here, $y = ax + b$ defines a supporting plane of the epigraph of $\psi$ at $x_0$). Let $x_0 = \int_X fdµ$, then we have $$\psi(\int_Xf d\mu) = \psi(x_0) = ax_0+b=a\int_Xf\mu + b = \int(af+b)d\mu \le \int(\psi\circ f)d\mu$$, q. e. d.

Generally I have two questions. First is how could the $a, b$ be guaranteed to exist? The other is if the $\mu$ is not from probability space and just Lebesgue measure or even general measure, will this Jensen inequality still hold (I think the requirement is $\mu$ should be finite measure at least)?

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  • $\begingroup$ 1) $\psi$ is convex and thus the epigraph also. 2) you need to rescale the inequality. It makes no sense if $\mu$ is infinite. $\endgroup$ – user251257 Feb 19 '16 at 4:51
  • $\begingroup$ @user251257: reply to 1. supporting hyperplane for convex set always exist and since this is $\mathbb R^2$, the hyperplane is just a line? 2. what did you mean by 'rescale'? $\endgroup$ – Bear and bunny Feb 19 '16 at 4:55
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    $\begingroup$ If $\mu$ is finite and non-zero, you could rescale it to a probability measure... $\endgroup$ – user251257 Feb 19 '16 at 4:57
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A counterexample with infinite measures. Let $m$ be Lebesgue measure on $[1,\infty)$, $f(x)=x^{-1}$, and $\varphi(x)=x^2$. Then

$$\varphi\left(\int fdm\right)=\infty \text{ and } \int \varphi\circ f dm=1.$$

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  • $\begingroup$ Yes. I think infinite measure make no sense. How about all finite general measure? should it work? $\endgroup$ – Bear and bunny Feb 19 '16 at 4:57
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    $\begingroup$ Yes. In this case for convex $\varphi$ $$\varphi\left(\frac{1}{\mu(X)}\int fd\mu\right)\le \frac{1}{\mu(X)}\int \varphi\circ fd\mu$$ $\endgroup$ – d.k.o. Feb 19 '16 at 5:00

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